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Digital Communication Subject Code: 17535 I m p o r t a n t I n
s t r u c t i o n s t o e x a m i n e r s : 1) The answers should
be examined by key words and not as word-to-word as given in the
model answer
scheme. 2) The model answer and the answer written by candidate
may vary but the examiner may try to assess the
understanding level of the candidate. 3) The language errors
such as grammatical, spelling errors should not be given more
Importance (Not
applicable for subject English and Communication Skills. 4)
While assessing figures, examiner may give credit for principal
components indicated in the figure. The
figures drawn by candidate and model answer may vary. The
examiner may give credit for anyequivalent figure drawn.
5) Credits may be given step wise for numerical problems. In
some cases, the assumed constant values may vary and there may be
some differenc
In case of some questions credit may be given by judgement on
part of examiner of relevant answer
7) For programming language papers, credit may be given to any
other program based on equivalent concept.
Q. No.
Sub Q.N. Answer
Marking Scheme
Q.1 Attempt any THREE of the following : 12-Total Marks
i) Define channel capacity with mathematical expression. 4M
Ans: (Definition - 2M, Mathematical expression -2M) Channel
capacity:- It is defined as the maximum rate at which data can be
transferred over the channel with an arbitrary small probability of
error. The unit of channel capacity is bits/sec.
Mathematical expression:- Channel capacity of a channel with
bandwidth B and band limited additive Gaussian white noise is given
by:-
C = B log 2 [1 + S/N] bits/sec where B channel bandwidth (Hz) S
Average signal power N Average noise power within the channel
bandwidth C Channel capacity bits/sec S/N ratio of total signal
power to total random noise power at the input of the receiver
Within the frequency limits of this channel i.e. over the
bandwidth.
ii) State sampling theorem and list its types. 4M
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Ans: (Statement - 3M, Types - 1M)
SAMPLING THEOREM: Sampling theorem states that a band-limited
signal of finite energy having the highest frequency component fm
Hz can be represented and recovered completely from a set of
Samples taken at a rate of fs samples per second provided that .
Here fs is the sampling Frequency. Types of Sampling:-
1. Natural sampling 2. Flat top Sampling
iii) Describe the need of multiplexing. 4M Ans: Need of
Multiplexing:-
Multiplexing is the set of techniques that allows the
simultaneous transmission of multiple signals across a signals data
link.
sends simultaneously i.e. multiplexer converts many into one, so
that at the receiving end also all input we get simultaneously.
Sending many signals separately is expensive and requires more
wires to send. So there is a need of multiplexing.
Efficient utilization of channel capacity. For example in cable
T.V distributor sends many channels through single wire.
4M
iv) List four applications of spread specturm modulation. 4M
Ans: (Each application - 1M)
Applications Of Spread Specturm Modulation:- The Spread Specturm
Communications are widely used today for Military, Industrial,
Avionics, Scientific, and Civil uses. The applications include the
following:
1. Jam-resistant communication systems 2. CDMA radios 3. High
Resolution Ranging: Spread Specturm Communications is often used in
high
resolution ranging. It is possible to locate an object with good
accuracy using SS techniques. for example where it could be used is
Global Positioning System (GPS).
4. WLAN: Wireless LAN (Local Area Networks) widely use spread
spectrum communications.
i. Infrared (IR) Communications ii. Direct Sequence Spread
Spectrum Communications
iii. Frequency Hopping Spread Spectrum Communications. 5.
Cordless Phones 6. Long-range wireless phones for home and industry
7. Cellular base stations interconnection. 8. Bluetooth. 9.
Satellite communication
4M
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b Attempt any one of the following : 6-Total Marks
(i) Draw the block diagram of basic digital communication system
and explain its working. 6M Ans:
Block diagram of basic digital communication system:
Explanation:- DISCRETE INFORMATION SOURCE:
The information to be transmitted originates here. These
information/messages may be available in digital form or it may be
available in an analog form.
If it is analog it is sampled and digitized using an A/D
converter to make the final source output to be digital in
form.
SOURCE ENCODER : The bit stream at the source output will have
considerable redundancy and so will not
be efficient representation of the message or information given
by the source, from the point of view of the number of digits
used.
A fewer number of digits might be sufficient to convey the
information. The source encoder therefore reduces the redundancy by
performing a one to one mapping of its input bit stream in to
another bit stream at its output, but with fewer digits.
Thus in a way it performs data compression. CHANNEL ENCODER:
The channel encoder is intended to introduce controlled
redundancy into the bit stream at its input in order to provide
some amount of error- correction capability to the data being
transmitted.
The data gets corrupted by the additive noise on the channel and
this gives rise to the possibility of the channel decoder
committing mistakes in the decoding of the data received from the
channel.
Redundancy helps in detecting erroneously decoded bits and makes
it possible to correct the errors before passing on the data to the
source decoder.
2M 4M
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DIGITAL MODULATOR: The physical channels are basically analog in
nature; the digital modulator takes each
digital binary digit at its input and maps it, in a one to one
fashion, into a continuous waveform.
PHYSICAL CHANNEL: The digitally modulated signal is passed on to
the physical channel, which is nothing
but the physical medium through which the signals are
transmitted. It may take a variety of forms- a pair of twisted
wires, coaxial cable, a wave guide, a
microwave radio, or an optical fiber. During its passage through
the channel, the signal gets corrupted by noise. This noise
may be thermal noise originating from electronic circuits or
atmospheric noise, or manmade noise, or as is generally the case, a
combination of most or all of them.
THE DIGITAL DEMODULATOR: The digital demodulator of the receiver
receives the noise corrupted sequence of
waveforms from the channel and by inverse mapping tries to give
at its output, an estimate of the sequence of the binary digits
that were available at the input of the digital modulator at the
transmitting end.
THE CHANNEL DECODER: The output sequences of digits from the
digital demodulator are fed to the channel
decoder. Using its knowledge of the type of coding performed by
the channel encoder at the transmitting end and using the
redundancy introduced by the channel encoder, it produces as its
output, the output of the source coder of the transmitter with as
few errors as possible.
THE SOURCE DECODER: Using its knowledge of the type of encoding
performed by the source encoder of the
transmitter, the source decoder of the receiver tries to
reproduce at its output, a replica of the output of the digital
source at the transmitting end.
ii) Explain Hamming distance (dmin.). How many errors can be
corrected and detected for the given minimum distance. 6M
Ans: (Hamming distance with example - 3M, Errors can be detected
and corrected with example - 3M)
The Hamming distance between two words is the number of
differences between corresponding bits. The minimum Hamming
distance is the smallest Hamming distance between all possible
pairs in a set of words. We first find all the Hamming
distances.
The dmin in this case is 3.
6M
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Errors can be detected and corrected with example To guarantee
the detection of up to s errors in all cases, the minimum Hamming
distance in a block code must be dmin = s+1. For dmin=3, no. of
bits error can be detected=dmin-1=3-1=2 To guarantee correction of
up to t errors in all cases, the minimum Hamming distance in a
block code must be dmin = 2t + 1. For dmin=3, no. of bits error can
be corrected=(dmin-1)/2=(3-1)/2=1
Q.2 Attempt any Two of the following : 16
a) Draw the block diagram of PCM transmitter and Receiver system
and explain function of each block. 8M
Ans: (PCM Transmitter & Receiver diagram = 4M, Explanation =
4M)
OR
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Block diagram of PCM transmitter:-
The analog signal x (t) is passed through a LPF (anti-aliasing
filter). The LPF band-limits the signal to fm band-limiting is
necessary to avoid the aliasing effect in the sampling process.
The pulse generator generates a train of pulses at a frequency
of fs such that fs > 2 fm. Thus, the Nyquist criterion is
satisfied.
The sampler block carries out flat-top sampling process on the
modulating signal at adequately high frequency. Then these samples
are subjected to the operation called Quantization in the
Quantizer.
The quantization process is the process of approximation of the
sampled signal. It assigns a particular level to which the sampled
value is near to.
The quantized PAM pulses are applied to an encoder. The encoder
converts each quantized level into an N-bit digital word (binary
pattern) such that Q = 2N where Q is the total number of
quantization levels.
The combination of the Quantizer and the Encoder is called as an
Analog-to-Digital Converter (A/D Converter). Thus, the signal
transmitted over the communication channel is a digitally-encoded
signal.
PCM RECEIVER: Block diagram of PCM Receiver
When the digitally-encoded signal is received at the receiver,
the receiver makes a
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Decision is made during each time slot, preferably at the center
of the time slot, compare the sample amplitude with a fixed
pre-
threshold.
Once it is decided that it is a 1 in a time slot, local pulse
generator in the receiver is triggered and it gives a clean, noise
free rectangular pulse. Although the received pulses were distorted
and corrupted by additive noise a clean pulse is locally
generated
-regeneration are then fed to a serial to parallel converter
then to a decoder which converts each code word into the
corresponding quantized sample value.
These samples are then passed through a low pass reconstruction
filter, which reconstructs an analog signal from these samples.
This analog signal will be an approximation to the original analog
message signal that was transmitted.
b) Illustrate BFSK signal generation with block diagram and
waveforms. State bandwidth requirement and draw its frequency
spectrum. 8M
Ans: (Block diagram = 2M, Explanation= 2M,Waveform =1M,
Bandwidth = 1M, Frequency Spectrum = 2M ) Block diagram of BFSK
signal Generation:-
Explanation:- In FSK, the frequency of the carrier is changed
with respect to the input bits 1 & 0. In case of binary data,
two carrier frequencies are used. The carrier frequency
corresponding to logic 0 or binary 0 is called as space
frequency and the carrier frequency corresponding to binary 1 is
called as mark frequency.
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FSK TRANSMITTER:
In general, for binary FSK, the carriers can be represented
by:
For binary 0, V0(t) = cos 0t = 0t
For binary 1, V1(t) = cos 1t = 1t Assume that high frequency is
transmitted for 1 and low frequency is transmitted for 0 Low
frequency = f0 = fc - High frequency = f1 = fc + The two carriers
may be generated from separate oscillators independent of one
another as
shown in the above figure. As shown in figure the input binary
data is given directly to the multiplier and is inverted
and given to second multiplier. Two different carriers have
different frequency generated by the two oscillators and
applied to the multipliers. The output of both the multipliers
is an ASK signal which is added by the summer. Thus,
the output of the adder is the FSK wave. Since there are two
different oscillators used for the carrier signal the combined
signal at
the summer therefore has discontinuous amplitude and phase which
is undesirable in FSK. Hence, we need a common carrier for FSK
modulation to avoid these discontinuities in amplitude and
phase.
Waveform:-
Frequency Spectrum:-
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Bandwidth Requirement:- Minimum bandwidth necessary for FSK is
given by:
B = (f1 + fb) (f0 fb) i.e. = (f1 f0) + 2fb Hence, B = 2fb+ 2fb =
4 fb
c) Explain working of CDMA. State its advantages and
disadvantages. 8M Ans: (Diagram = 2M , Explanation =2M, Advantages
= 2M, Disadvantages =2M)
CDMA is used along with spread spectrum modulation technique,
which neither uses frequency channels nor time slots.
In SSM (spread spectrum modulation) spreading is achieved by PN
code. If such unique code is assigned to each individual user, the
demodulation will be possible only if the code matches at the
receiver end. This enables the multiple access.
CDMA system uses same frequency band and transmit
simultaneously. They can use the whole available bandwidth for all
the time. The transmitted signal is recovered by co-relating the
received signal with the PN code used by the transmitter.
CDMA property 1. Non-interference with existing system 2.
Anti-jam and interference rejection 3. Information security 4.
Accurate ranging 5. Multipath tolerance CDMA allows all the users
to occupy all channels at the same time. Transmitted signal
is spread over the whole band and each voice or data call is
assigned a unique code to differentiate it from other calls carried
over the space spectrum.
All the users in CDMA use same carrier and may transmit
simultaneously. Each user has its own pseudorandom code word which
is orthogonal to all other code words. For detection of message
signal the receiver needs to know the code word use by transmitter.
Each user operates independently with no. of knowledge of other
users.
CDMA MULTIPLE ACCESS USING DS-SPREAD SPECTRUM
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CDMA with FHSS:- As in DS spread spectrum multiple access is
achieved in FHSS also by assigning a unique
PN code to each user, which in this case controls the
frequency-hopping pattern. These codes that are assigned must be so
chosen that collision do not occur. The frequency produced by the
frequency synthesizer during a chip period depends on the
PN sequence value during that chip period. So sometimes it may
so happen that two or more users have at a given time, the same
PN
sequence values produced by their respective PN sequence
generators. In that case a collision is said to have occurred in
the spectrum.
Whether it is slow hopping or fast hopping FHSS, when a
collision occurs it results in considerable increase in detection
errors.
Advantages:- (Any 2)
The CDMA does not require any synchronization.
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It has more number of users can share the same bandwidth. It is
well-matched with other cellular technologies. Due to code word
allocated to each user, interference is reduced. Efficient
practical utilization of fixed frequency spectrum.
Disadvantages:- (Any 2) The system is more complicated. Guard
band and guard time both are required to be provided. As the number
of users increases, the overall quality of services decreases.
Q.3 A) Attempt any FOUR of the following : 16
a) Explain Companding with neat diagram. 4M Ans: (Diagram = 2M ,
Explanation =2M)
Diagram Of COMPANDING :-
Explanation:- The combination of compressor and expander is
known as compander which performs the Companding process. It is
used to increase the signal to quantization error ratio for weak
signal. Compression process provides high gain to weak signals
& less gain to strong signals. Practically implemented using a
device of logarithmic (log) characteristics. Here weak signals are
artificially boosted to improve signal to quantization noise ratio.
This process also attenuates strong signals, keeping the value of
S/NQ almost constant. It takes place before the actual quantization
process, in both the 1st & 2nd quadrants.
Expanding process provides a higher gain to strong signals &
less gain to weaker signals. Practically implemented using a device
of exponential (antilog) characteristics. This is because weak
signals were artificially boosted & hence need to be restored
back. This process also magnifies strong signals, keeping the value
of S/NQ almost constant. It takes place after the actual
quantization process, towards the receiver side. Compression at the
transmitter side. Expansion at the Receiver side. At the
transmitter end the information signal is passed through compressor
where the
2M 2M
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signal is amplified more at low amplitude than at high
amplitude. At the receiver side, an inverse operation is performed
to recover the original information signal. This is achieved by an
expander. The Companding Characteristics:-
. b) Draw the block schematic of DM transmitter and Receiver. 4M
Ans:
DM Transmitter Block Diagram:-
DM Receiver Block Diagram:-
2M 2M
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c) Describe synchronous TDM with block diagram transmitter. 4M
Ans: (Diagram = 2M , Explanation =2M)
Diagram Of Synchronous TDM Transmitter:-
Explanation:- In synchronous time division multiplexing, each
device (transmitter) is allotted with a fixed time slot, regardless
of the fact that the device (transmitter) has any data to transmit
or not. The device has to transmit data within this time slot. If
the device (transmitter) does not have any data to send then its
time slot remains empty. As shown in the below figure, the various
time slots are arranged into frames and each frame consists of one
or more time slots dedicated to each device (transmitter). For
example, if there are 3 devices, there will be 3 slots in each
frame. Similarly, if there are 5 devices, there will be 5 slots in
each frame. The above figure shows 4 devices (transmitter A,
transmitter B, transmitter C, and transmitter D) that have 4
dedicated time slots (time slot A, time slot B, time slot C and
time slot D). The transmitter A data is sent at time slot A,
transmitter B data is sent at time slot B, transmitter C data is
sent at time slot C and transmitter D data is sent at time slot D.
In the time frame 2, the transmitter B and C does not have any data
to send so the time slot B and C remains empty. The main drawback
of synchronous time division multiplexing is that the channel
capacity is not fully utilized. Hence, the bandwidth goes
wasted.
2M 2M
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d) Compare QPSK QASK (any four points). 4M Ans: Parameter QPSK
QASK
1.Information is transmitted by change in
phase Amplitude and phase
2. Number of bits/symbol
N=2 N=3 Or 4 Or 5 & so on
3.Number of possible symbols
Four M=2N
4.Detection method coherent coherent 5.minimum bandwidth fb
2fb/N 6.symbol duration 2Tb NTb 7.Noise immunity Comparatively
less than QAM Better than QPSK
8.System complexity Less complex than QAM
More complex than QPSK
9.Probability of error More than QAM Less than QPSK
10.Performance of system
Less than QAM Better than QPSK
Each Point1M
e) Describe QAM transmitter with block diagram. 4M Ans: QAM
transmitter block diagram:-
QAM transmitter Explanation:- The bi stream b(t) is applied to
the serial to parallel converter, operating on a clock which has
period of Ts, Which is the symbol duration. The bits b(t) are
stored by the converter and then presented in the parallel form.
The four bit symbols are bk +3, bk +1, and bk. Out of these four
bits the first two bits are applied to a D/A converter and the
other two bits are applied to the second D/A converter. The Output
of the first converter is Ae(t),which is modulated by the carrier T
whereas is the output of the second D/A converter Ao is modulated
by the carrier sin
Ae(t) Ao(t) are voltages levels generated by the converter -3
-1, +1,+3 volts The balanced modulator outputs are added together
to get the QASK output signal which is expressed as: QASK(t) =
Ae(t) ct + Ao(t) ct
2M 2M
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Q.4 a) Attempt any THREE of the following: 12
(i) State advantages and disadvantages of digital communication
system. 4M Ans: (Any 4 advantages = 2M , Any 2 disadvantages
=2M)
Advantages of digital communication System: 1. Digital systems
are simple and easy to build. 2. Insensitive to variations in
atmospheric conditions like temperature, humidity etc. so
highly robust. 3. Storage and retrieval of voice, data or video
is easy& inexpensive. 4. It offers considerable flexibility as
voice, data, video can all be multiplexed using
TDM, signal and image processing operations like compression of
voice and image signal can be easily used etc
5. Cost of digital communication systems are coming down because
of improvement of
6. Error correction codes ensure fairly good protection against
noise and interferences. 7. Powerful encryption and decryption
algorithms are available for digital data so as to
maintain high level of secrecy of communication. 8. In digital
communication system, repeaters are used as regenerators so signal
reaching
at the destination can be almost error free. So this system can
be used for long-haul communication.
9. Multiple data can be send simultaneously using multiplexing.
Disadvantages of Digital Communication System:
1. The transmission of digitally encoded analog signals requires
significantly more bandwidth.
2. Digital transmission requires precise time synchronization
between the clocks in the transmitter and receiver.
3. Digital transmission systems are incompatible with older
analog transmission systems. 4. Digital communication system
requires greater bandwidth
2M 2M
(ii) Explain the need of using adaptive delta modulation. 4M
Ans: Necessity of adaptive delta modulation technique:-
In delta modulation, the step size is constant so that its slope
overload distortion and granular noise both cannot be controlled.
These drawbacks can be controlled by using adaptive delta
modulation wherein the step size is variable. Thus with adaptive
delta modulation the following are the advantages- 1. Slope
overload distortion and granular noise problem in is reduced. 2.
Improved signal to noise ratio. 3. Wide dynamic range is achieved
with variable step size. 4.Better bandwidth utilization than delta
modulation
4M
(iii) Draw and explain spread spectrum modulation system. 4M
Ans:
(Diagram = 2M , Explanation =2M
Model Of a Spread Specturm modulation System:-
2M
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Explanation Of Spread Spectrum Modulation System:-
1. Basic elements of a spread spectrum signal modulation system
is shown below. 2. Channel encoder adds extra bits to the
information binary sequence for error
detection & correction purpose.
3. PN sequence generation at the transmitter & receiver
generates identical PN binary valued sequence
4. PN sequence is impressed on the information signal at the
modulator (Tx) and remove from the received signal at the
Demodulator.
5. Synchronization of the PN sequence generator at the receiver
with the PN sequence contained in the incoming received signal is
required in order to demodulate the received signal.
6. Prior to the transmission of information Synchronization may
be achieved by transmitting a fixed PN sequence pattern which the
receiver will recognize in the presence of interference with high
probability.
2M
(iv) For the binary data system 10110010 draw Unipolar RZ, Polar
RZ, split phase Manchester and AMI. 4M
Ans:
1M each wave-form
b) Attempt any ONE of the following 6M (i) Explain working of
CRC generator and checker. 6M
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Ans: (Diagram = 2M , Explanation =2M and Example = 2 M)
Explanation Of Cyclic Redundancy Check (CRC):- With CRC the entire
data stream is treated as long continuous binary number. In this
method, a sequence of redundant bits, called the CRC or the CRC
remainder, is appended to the end of the unit so that the resulting
data unit becomes exactly divisible by a second, predetermined
binary number. At its destination, the incoming data unit is
divided by the same number. If at this step there is no remainder,
the data unit assume to be correct and is accepted, otherwise it
indicate that data unit has been damaged in transmission and
therefore must be rejected The redundancies bits are used by CRC
are derived by dividing the data unit by a predetermined divisor.
The remainder is the CRC. Diagram Of Cyclic Redundancy Check:-
Example:
4M 2M
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(ii) Explain frequency hopping. Compare slow and fast frequency
hopping (four points) 6M Ans: (Explanation =2M and each point of
comparison =1 M)
FHSS: It combines spread spectrum modulation with MFSK.It is the
process of modifying frequency of MFSK signal using frequency hope
generated by bits of PN sequence. Different types of frequency
hopping:-Slow and fast frequency hopping. FHSS(Frequency Hoping
Spread spectrum):
Signal is broadcasting over seemingly random series of
frequencies Receiver hops between frequencies in sync with
transmitter. Eavesdroppers hear unintelligible blips. Jamming on
one frequency affects only a few bits.
Slow frequency hopping Fast frequency hopping 1.More than one
symbols are transmitted per frequency hop
1.More than one frequency hops are required to transmit one
symbol
2.chip rate is equal to symbol rate 2.chip rate is equal to hop
rate 3.symbol rate is higher than hop rate 3.hop rate is higher
than symbol rate 4.same carrier frequency is used to transmit one
or more symbols
4.one symbol is transmitted over multiple carrier in different
hops
5.A jammer can detect this signal if the carrier frequency in
one hop is known
5.A jammer cannot detect this signal because one symbol is
transmitted using more than one carrier frequencies
2M 1M each point
Q.5 Attempt any TWO of the following: 16
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a) Draw block diagram of FDM transmitter and Receiver and
explain function of each block.
8M
Ans:
(block diagram of Tx/Rx = 2M each , Explanation =2M each ) FDM
TRANSMITTER:
In FDM, the transmission channel is shared by multiple signals,
each being allotted a portion of the spectrum of the bandwidth. A
generalized block diagram of FDM transmitter is depicted in
Figure
FDM Transmitter
A number of analog signals (or digital signals converted into
analog) mi(t), i = 1, 2, . . ., n are multiplexed onto the same
transmission medium. Each signal mi(t) is modulated onto a carrier
fci.
As multiple carriers are to be used, each is referred to as a
subcarrier. Any type of analog modulation may be used. The
resulting analog signals are summed together to produce a composite
signal, mb(t).
The composite signal may be shifted, as a whole, to another
carrier frequency by an additional modulation step.
This second modulation step need not use the same modulation
technique as the first. Thus, the FDM signal generated may be
transmitted over a suitable medium.
FDM RECEIVER:
2M 2M 2M
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FDM Receiver
The FDM receiver is shown in Figure. The FDM signal is received
by the receiver and
demodulated to retrieve the composite signal mb(t) which is
further amplified. This composite signal is passed through nband
pass filters, each filter having a center
frequency equal to the subcarrier frequency fci. In this way the
composite signal is split into its component signals. Each
component
signal is further demodulated to obtain the analog outputs that
were originally transmitted. If required, these signals are stored
and displayed.
2M
b) Draw and explain block diagram of DPSK transmitter and
receiver. 8M Ans: (block diagram of Tx/Rx = 2M each , Explanation
=2M each )
DPSK MODULATOR:
The generation block diagram of DPSK signal is shown in Figure
3.16. The data stream to be transmitted, d(t), is applied to one
input of an exclusive-OR logic gate. To the other gate input the
output of the exclusive-OR gate b(t) delayed by time Tb allocated
to one bit is applied. This second input is then b(t - Tb).
Figure : DPSK Transmitter
2M
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The truth table of the exclusive-OR gate is shown below:
d(t) b(t - Tb) b(t) Logic Level Voltage Logic Level Voltage
Logic Level Voltage
0 -1 0 -1 0 -1
0 -1 1 1 1 1 1 1 0 -1 1 1 1 1 1 1 0 -1
From figure 3.16, b(t) is given by, b(t) = d(t) b(t Tb)
Input Data d(t) 1 0 1 1 1 0
Delayed input b(t Tb) 0 1 1 0 1 0
XOR Output b(t) 0 1 1 0 1 0 0
Output Phase 0° 0° 180° 0° 180° 180°
DPSK input (at Receiver) 180° 0° 0° 180° 0° 180° 180°
Recovered data stream 1 0 1 1 1 0
It is observed that when d(t) = 0, b(t) = b(t Tb) and when d(t)
= 1, b(t) = b(t Tb). As
seen in Figure 3.16,b(t) is applied to a level shifter which
assigns a positive voltage level when b(t) = 1 and a negative
voltage level when b(t) = 0. The level shifter output is then
applied to a balanced modulator to which a carrier signal scos ct
is also applied. The modulator output, which is the transmitted
signal is given by,
VDPSK ct
= (±1) ct DPSK DEMODULATOR:
The DPSK demodulator is shown in Figure. Here the received
signal and the received signal delayed by the bit time Tb are
applied to a balanced modulator. The balanced modualtor output is
given by,
b(t)b(t Tb)(2Ps c c(t Tb)
The output of the balanced modulator is applied to the
integrator which suppresses the double frequency term. The first
term [ b(t)b(t Tb)Ps cTb]is the required signal and cTb is selected
in such a way so that cTb where n is an integer) so that
cTb= + 1 and the signal output will be as large as possible.
Further, with this selection, the bit duration encompasses an
integral number of clock cycles and the integral of the double
frequency term is exactly zero.
2M
2M
2M
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Output of the integrator = b(t) b(t - Tb) Ps Tb
DPSK Receiver
The transmitted data bit d(t) can be determined from the product
b(t)b(t Tb). If there is no phase change between b(t) and b(t Tb)
then d(t) = 0.i.e. if b(t) = b(t - Tb).In this case b (t) b (t Tb)
= + 1. No phase change between b (t) and b (t Tb) means both are 1
or both are 0, which means both are +1 or both are -1 hence the
multiplication is always positive.
If there is a phase change between them, then either b(t)or b(t
Tb) = - 1. i.e. if b(t) b(t - Tb) In either case, b(t)b(t Tb) = -
1. Then d(t) = 1. Phase change between b (t) and b (t Tb) means
either is 1 and other is 0, which means one is +1 and other is -1
hence the multiplication is always negative.
If the comparator receives +ve voltage bit received is 0 , if it
receives ve voltage bit received is 1.
c) Describe DSSS transmitter and receiver working with block
diagram. 8M Ans: (block diagram of Tx/Rx = 2M each , Working =2M
each )
DSSS transmitter and receiver working with block diagram:-
Explanation:- The information signal undergoes primary
modulation by PSK, FSK or other narrow band
4M
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modulation and secondary modulation with spread spectrum
modulation. Spread spectra are obtained by multiplying the primary
modulated signal and the square wave, called the PN sequence.
Contrariwise, as with commercial radio, there are cases where
spread modulation is applied to the data first, and narrow band
modulation such as PSK or FSK is applied afterwards. The figure
below is an example of spread spectrum modulation and demodulation
using PSK for primary modulation. Receiver: If despreading is
applied to the received diffuse wave, it returns to the PSK or FSK
modulated wave resulting from primary modulation. Then, as with
narrowband demodulation, if the despread wave and local signal are
multiplied, and appropriate low pass processing is applied, the
information signal is obtained. Despreading involves multiplying
the same PN code as that used at the transmitting ennecessary to
synchronize the receiving wave and PN code. There are two
processing methods on the receiving side, demodulation of the
information signal after despreading, and obtaining a positive and
negative PN code by multiplying the local signal by the receiving
wave and despreading using correlation detection. With the former
there is process gain but the problem of synchronization remains.
With the latter, the spectrum density of the receiving wave itself
is low, and regeneration of the local carrier for performing
synchronous detection is a problem. Commercial SS radio equipment
uses the latter, but it requires considerable power and has a short
communication range. Despreading: The signal that enters the
antenna of the receiver includes outside interference waves and
noise. If this signal is despread, the signal component returns to
a narrowband modulated wave and the interference components are
diffused, expanding the spectrum infinitely so that its power
density falls. Therefore, by inputting the signal with frequency
band restricted using a BPF, the interference component power that
falls into the demodulation frequency band is reduced. The
occurrence of errors is calculated using a stochastic process, so
ultimately, using a spread spectrum results in fewer errors, and
this is why spread spectrum communication is resistant to
interference. Demodulation: Demodulation is normal narrowband
demodulation. The local signal is regenerated from the receiving
wave and after multiplication by the receiving wave, unnecessary
components are eliminated with an LPF. Primary modulation uses PSK,
so synchronous detection is necessary. PNsequence : The PN sequence
is switched at a far faster speed than the symbol rate of the
information signal and its spectrum covers a wide band. For this
reason, the spectrum of the modulated
PN sequences must meet the conditions required for spread
spectrum modulation such as the relationship of the numbers 1 and
0.
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Q.6 Attempt any FOUR of the following: 16
a) Compare analog and digital pulse modulation. 4M Ans: SR.N
O PARAMETER ANALOG PULSE
MODULATIONDIGITAL PULSE MODULATION
1 Nature of transmittedsignal
Pulse with varyingparameters(amplitude, width or position of the
pulse)
Digital signal i.e in the
2 Noise immunity Poor Excellent 3 Bandwidth
requirement Lower then digital Higher due to higher bit
rate 4 Multiplexing used FDM/TDM TDM
5 Types PAM,PPM,PWM DM,ADM,PCM,DPCM
1M for each point
b) Compare FDMA and TDMA (four points) 4M Ans: Sr No
PARAMETER
FDMA TDMA
1
Definition
Entire band of frequencies is divided into multiple RF
channels/carriers. Each carrier is allocated to different
users.
Entire bandwidth is shared among different subscribers at fixed
predetermined or dynamically assigned time intervals/slots.
2
Bandwidth available
Overall bandwidth is shared among many stations. Less BW is
available.
Time sharing of satellite transponder takes Place. Hence large
BW is available.
3 Synchronization
Synchronization is notnecessary
Synchronization is essential
Due to nonlinearity of devices Intermodulation products are
generated due to interference
Due to incorrect synchronization there can be
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4 Interferencebetween adjacent channels.
interference between the adjacent time slots.
5 Handoff Hard handoff Hard handoff
6
Application
GSM , PDC(pacific digital cellular), Radio, TV
Advanced mobile phone, system(AMPS), Cordless telephone
c) Draw and explain the block diagram of ASK with suitable
diagram. 4M
Ans:
Explanation of ASK Transmitter The ASK technique of binary
modulation is illustrated in Figure where modulating
signal consists of unipolar pulses. Because in this case the
carrier is switched ON and OFF, this method is also known
as ON-OFF keying. For the entire time the binary input is high,
the output is a constant amplitude,
constant frequency signal and for the entire time the binary
input is low, the carrier is off.
Ps is signal power given by (Amplitude) 2 /2 ASK is given by:
ASK(t) = b(t) ct
2M 2M
d) bandwidth 3100 Hz and signal to noise ratio 20 dB. 4M
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Ans:
4M
e) Explain BPSK signal generation with block diagram and
waveform. 4M Ans: The generation block diagram of BPSK is shown
below
BPSK SIGNAL GENERATOR
The unipolar binary input is converted into a bipolar signal
through a level converter. The two dc levels generated by the
binary comparator (level converter) alters the flow
of current through the balanced modulator resulting in the phase
change of the carrier generated by the local oscillator (reference
carrier generator) between 0° and 180°.
2M 1M
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The BPF is required in order to maintain channel bandwidth
restriction and to reduce ISI (Inter-symbol Interference).
If the data = 1, voltage level = +1, hence the o/p will be = +1x
ct If the data = 0, voltage level = -1, hence the o/p will be = -1x
ct
(- ct)= ( ct +1800) Mathematically, BPSK is given by:
For binary 1, VBPSK(t) = ct
For binary 0, VBPSK(t) = ct+ 180°)
BPSK WAVEFORMS
1M